What is the relation between the definitions of Yangian $Y(sl_2)$ and quantum affine algebra $U_q(\widehat{sl_2})$?
There are two definitions of $U_q(\widehat{sl_2})$. The following is Jimbo presentation. The quantum affine algebra $U_q(\widehat{\mathfrak{sl}_2})$ is an associated algebra generated by $e_0^{\pm}, e_1^{\pm}, K_0^{\pm 1}, K_1^{\pm 1}$ subject to the relations: \begin{align} & K_0 K_1 = K_1 K_0, \\ & K_i K_i^{-1} = K_i^{-1} K_i = 1, \ i=0,1, \\ & K_i e_i^{\pm} = q^{\pm 2} e_i^{\pm} K_i, \ i=0,1, \\ & K_i e_j^{\pm} = q^{\mp 2}e_j^{\pm} K_i, \\ & e_0^{\pm} e_1^{\mp} = e_1^{\mp} e_0^{\pm} \\ & e_i^{+} e_i^{-} - e_i^{-}e_i^{+} = \frac{K_i-K_i^{-1}}{q-q^{-1}}, \ i=0,1, \end{align} and quantized Serre relations: \begin{align} (e_i^{\pm})^3 e_j^{\pm} - [3] (e_i^{\pm})^2 e_j^{\pm}e_i^{\pm} + [3]e_i^{\pm} e_j^{\pm} (e_i^{\pm})^{2} -e_j^{\pm}(e_i^{\pm})^3 = 0, \ i \neq j. \end{align}
The following is Drinfeld presentation. The quantum affine algebra $U_q(\widehat{\mathfrak{sl}_2})$ is an associated algebra generated by $x_m^{\pm}, h_r, K^{\pm 1}, c^{\pm 1}$, $m \in \mathbb{Z}, r \in \mathbb{Z}-\{0\}$, subject to the relations: \begin{align} & c^{\pm 1} \text{ are in the center}, \\ & K K^{-1} = K^{-1} K = 1, \\ & c c^{-1} = c^{-1} c = 1, \\ & [K, h_r] = 0, \\ & K x^{\pm} = q^{\pm 2} x_i^{\pm} K, \\ & [h_k, x_l^{\pm}] = \pm \frac{1}{k}[2k]c^{\mp |k|} x_{k+l}^{\pm}, \\ & x_{k+l}^{\pm} x_l^{\pm} - q^{\pm 2}x_l^{\pm} x_{k+l}^{\pm} = q^{\pm 2} x_k^{\pm} x_{l+1}^{\pm} - x_{l+1}^{\pm} x_k^{\pm}, \\ & [h_k, h_l] = \delta_{k,-l} \frac{1}{k} [2k] \frac{c^k-c^{-k}}{q-q^{-1}}, \\ & [x_k^+, x_l^-] = \frac{1}{q-q^{-1}} [c^{k-l}\psi_{k+l}-\phi_{k+l}], \end{align} where $\psi_k$, $\phi_k$ are given by: \begin{align} & \sum_{m=0}^{\infty} \psi_m z^m = K \exp\left( (q-q^{-1}) \sum_{s=1}^{\infty} h_s z^s \right), \\ & \sum_{m=0}^{\infty} \phi_{-m}z^{-m} = K^{-1} \exp\left( -(q-q^{-1}) \sum_{s=1}^{\infty} h_{-s} z^{-s} \right), \end{align} and $\psi_k=0$, $\phi_{-k}=0$, $k < 0$.
The definition of $Y(sl_2)$ is defined as follows.
Thank you very much.